Simplify.
$\sqrt{147}$
$\sqrt{147}$
Simplify.
$\sqrt{147}$
To simplify the radical \(\sqrt{147}\), we first analyze the radicand, which is 147. We look for its prime factors to identify any perfect square factors. A perfect square factor is a number of the form \(k^2\) where \(k\) is an integer. We test for small prime factors: 147 is not divisible by 2 (it's odd). Checking for 3: the sum of digits is \(1+4+7=12\), which is divisible by 3, so 147 is divisible by 3. Dividing 147 by 3 gives \(147 \div 3 = 49\). We recognize that 49 is a perfect square, specifically \(7 \times 7\) or \(7^2\). Therefore, the prime factorization of 147 is \(3 \times 7^2\).
Now that we have the prime factorization \(147 = 3 \times 7^2\), we can use the property of radicals which states that \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\) for non-negative real numbers \(a\) and \(b\). Applying this to our expression, we have:
\[\sqrt{147} = \sqrt{49 \times 3} = \sqrt{49} \times \sqrt{3}.\]
Since \(\sqrt{49} = 7\), we can substitute this back into our equation:
\[\sqrt{147} = 7\sqrt{3}.\]
This form is the simplified version of the radical because 3 has no perfect square factors other than 1.
Simplifying square roots is a foundational skill in geometry and physics. For example, if you are calculating the diagonal of a rectangular floor with sides \(\sqrt{49}\) meters and \(\sqrt{98}\) meters, you might end up with an expression like \(\sqrt{49+98} = \sqrt{147}\). Simplifying this to \(7\sqrt{3}\) meters allows you to estimate the distance more easily (since \(\sqrt{3} \approx 1.732\), the length is approximately \(7 \times 1.732 = 12.124\) meters) and helps in further algebraic manipulations involving the measurements.